LECTURE NOTES ON REAL INTERPOLATION

TERENCE TAO

1. Overview

The purpose of these notes is to prove the following interpolation theorem:

Theorem 1.1 (Marcinkeiwicz interpolation theorem). Let p_{0}, p_{1}, q_{0}, q_{1} be exponents in [1, ∞] such that p_{0} ≤ q_{0}, p_{1} ≤ q_{1}, and q_{0} ≠ q_{1}. Let (X, μ) and (Y, ν) be σfinite measure spaces, and let T be a sublinear operator which maps L^{p_{0}}(X)+L^{p_{1}}(X) to measurable functions in Y, thus we have |T (f + g)(y)| ≤ |T f(y)| + |T g(y)| for all f, g ∈ L^{p_{0}}(X) + L^{p_{1}}(X) and almost every y, and |T (cf)| = c|T f| for all c > 0. Suppose we also have the weak-type estimates

∥T f∥_{L^{q_{0},∞}(Y)} ≤ M_{0}∥f∥_{L^{p_{0}} (X)} for all f ∈ L^{p_{0}}(X)

and

∥T f∥_{L^{q_{1},∞}(Y)} ≤ M_{1}∥f∥_{L^{p_{1}} (X)} for all f ∈ L^{p_{1}}(X).

Then we have the strong-type estimate

∥T f∥_{L^{q_{θ}} (Y)} ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)} for all f ∈ L^{p_{θ}}(X)

for all 0 < θ < 1, where C is some quantity depending on p_{0}, p_{1}, q_{0}, q_{1}, θ and

\frac{1}{p_{θ}} = \frac{1 − θ}{p_{0}} + \frac{θ}{p_{1}} ; \frac{1}{q_{θ}} = \frac{1 − θ}{q_{0}} + \frac{θ}{q_{1}} .

We will use two tools to establish the theorem. The first is an upper bound on ∫_{F} |T f| for various sets F and special kinds of functions f, which uses the hypotheses of the theorem. The second is a rather general “wedding cake decomposition” which has many uses beyond this interpolation theorem. Combining both estimates in the “natural” way, we will then obtain the conclusion.

Let us first dispense with an easy case when p_{0} = p_{1}. In that case, all one needs to do is establish the inequality

∥g∥_{L^{q_{θ}} (Y)} ≤ C∥g∥_{L^{q_{0},∞}(Y)}^{1−θ}∥g∥_{L^{q_{1},∞}(Y)}^{θ}

for all g ∈ L^{q_{0},∞}(Y)∩L^{q_{1},∞}(Y), as the claim then follows by applying this inequality with g := T f. But this follows by considering the distribution function of g; we leave this as an exercise to the reader.

2. An estimate on |T f|

We begin with a preliminary estimate on T f.

Lemma 2.1. Let f be supported on a set E ⊆ X on a set of finite measure, and obey the bounds |f(x)| ≤ A on E. Let F ⊂ Y have finite measure.

• (i) There exists a set F′ ⊂ F of measure ν(F′) ≤ \frac{1}{2}ν(F), such that

∫_{F \F′} |T f| ≤ 2^{1/q_{θ}}AM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F)^{1/q′_{θ}}

for all 0 ≤ θ ≤ 1.

• (ii) For every 0 < θ < 1 we have

∫_{F} |T f| ≤ CAM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F)^{1/q′_{θ}}

where

C := \frac{2^{1/q_{θ}}}{1 − 2^{−1/q′_{θ}}} .

Remark: This lemma would follow easily from the Minkowski interpolation theorem and Hölder’s inequality, though of course we cannot yet use that interpolation theorem in our arguments.

Proof We first prove (i). Observe that it suffices to prove this inequality only for the extreme cases θ = 0 and θ = 1. This is because the expression 2^{1/q_{θ}}AM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F)^{1/q′_{θ}} is either increasing in θ, decreasing in θ, or constant. In any of these cases, the intermediate values of this quantity are greater than or equal to one of the endpoints. So if we know the result is true at both endpoints, it automatically implies the result in the interior as well.

So now we prove (i) for θ = 0 or θ = 1. The claim is trivial if F has zero measure, so assume F has positive measure. From hypothesis we have

∥T f∥_{L^{q_{θ} ,∞}(Y)} ≤ M_{θ}∥f∥_{L^{p_{θ}} (X)} ≤ M_{θ}∥A1_{E}∥_{L^{p_{θ}} (X)} ≤ AM_{θ}ν(E)^{1/p_{θ}}.

Thus we have

|ν(\{y ∈ Y : |T f(y)| ≥ α\}| ≤ \frac{A^{q_{θ}}M_{θ}^{p_{θ}}ν(E)^{q_{θ}/p_{θ}}}{α^{q_{θ}}}

for any α > 0. If we select α to solve the equation

\frac{A^{q_{θ}}M_{θ}^{p_{θ}}ν(E)^{q_{θ}/p_{θ}}}{α^{q_{θ}}} = \frac{1}{2}ν(F)

and then set

F′ := {y ∈ F : |T f(y)| ≥ α}|

then we clearly have ν(F′) ≤ \frac{1}{2}ν(F), and

∫_{F \F′} |T f| ≤ ∫_{F \F′} α ≤ αν(F).

Using the definition of α, and some algebra, we conclude that

αν(F) = 2^{1/q_{θ}}AM_{θ}μ(E)^{1/p_{θ}}ν(F)^{1/q′_{θ}}

and the claim follows.

Now we prove (ii). Set F_{0} := F. Applying (i), we can find a set F_{1} ⊆ F_{0} with ν(F_{1}) ≤ \frac{1}{2}ν(F_{0}) and

∫_{F_{0}\F_{1}} |T f| ≤ 2^{1/q_{θ}}AM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F_{0})^{1/q′_{θ}}

Applying (i) again, we can find F_{2} ⊆ F_{1} with ν(F_{2}) ≤ \frac{1}{2}ν(F_{1}) such that

∫_{F_{1}\F_{2}} |T f| ≤ 2^{1/q_{θ}}AM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F_{1})^{1/q′_{θ}}

and so on and so forth. More generally, we can find F_{0} ⊇ F_{1} ⊇ F_{2} ⊇ … such that ν(F_{n}) ≤ 2^{−n}ν(F_{0}) and

∫_{F_{n}\F_{n+1}} |T f| ≤ 2^{−n/q′_{θ}}2^{1/q_{θ}}AM_{0}^{1−θ}M_{1}^{θ}μ(E)^{1/p_{θ}}ν(F_{0})^{1/q′_{θ}}.

Observe that ⋂_{n} F_{n} has measure zero. We can then sum these integrals in n and use monotone convergence (and the geometric series formula to obtain the claim. Note that since 0 < θ < 1 and q_{0} ≠ q_{1}, we have 1 < q_{θ} < ∞ and hence 2^{−1/q′_{θ}} is strictly less than one.

3. The wedding cake decomposition

To use the above estimate on |T f|, we need the following decomposition, which is also useful in several other applications.

Lemma 3.1 (Wedding cake decomposition). Let f ∈ L^{p}(X) be a simple function for some 0 < p < ∞. Then there exists a decomposition f = ∑_{n=−∞}^{∞} f_{n}, with all but finitely many of the f_{n} being non-zero, such that for each n, there exists a set E_{n} ⊂ X such that 2^{n}1_{E_{n}} ≤ |f_{n}| ≤ 2^{n+1}1_{E_{n}}. Furthermore, we have the bound

∑_{n=−∞}^{∞} 2^{np}μ(E_{n}) ≤ ∥f∥_{L^{p}(X)}^{p}.

Proof We set

E_{n} := {x ∈ X : 2^{n} ≤ |f_{n}(x)| ≤ 2^{n+1}}

and

f_{n}(x) := 1_{E_{n}}(x)f(x).

The claims of the lemma are then easily verified. Note that

∥f∥_{L^{p}(X)}^{p} = ∫_{X} |f|^{p} = ∑_{n} ∫_{E_{n}} |f|^{p} ≥ ∑_{n} ∫_{E_{n}} 2^{np} = ∑_{n} 2^{np}μ(E_{n}).

We can now obtain the key estimate used to establish the interpolation theorem.

Proposition 3.2. Let 0 < θ < 1, and let f ∈ L^{p_{θ}}(X) and g ∈ L^{q′_{θ}}(Y) be simple functions. Then we have

∫_{Y} |T f||g| ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}∥g∥_{L^{q′_{θ}} (Y)}

for some C.

Proof We may assume that ∥f∥_{L^{p_{θ}} (X)}, ∥g∥_{L^{q′_{θ}} (Y)} are non-zero since the claim is easy otherwise. We can then normalize ∥f∥_{L^{p_{θ}} (X)} = ∥g∥_{L^{q′_{θ}} (Y)} = 1.

Since q_{0} ≠ q_{1} and q_{θ} is strictly between q_{0} and q_{1}, we see that 1 < q_{θ} < ∞ and hence 1 < q′_{θ} < ∞. Similarly, since p_{0} ≠ p_{1} we see that p_{θ} is also finite. Then we can apply the wedding cake decomposition to obtain

f = ∑_{n} f_{n}; 2^{n}1_{E_{n}} ≤ |f_{n}| ≤ 2^{n+1}1_{E_{n}}; ∑_{n} 2^{np_{θ}}μ(E_{n}) ≤ 1

and

g = ∑_{m} g_{m}; 2^{m}1_{F_{m}} ≤ |g_{m}| ≤ 2^{m+1}1_{F_{m}}; ∑_{m} 2^{mq′_{θ}}ν(F_{m}) ≤ 1.

By subadditivity, and the fact that only finitely many of the f_{n} and g_{m} are non-zero, we then have

∫_{Y} |T f||g| ≤ ∑_{n} ∑_{m} ∫_{Y} |T f_{n}||g_{m}| ≤ ∑_{n} ∑_{m} 2^{m+1} ∫_{F_{m}} |T f_{n}|.

Applying

To deal with this expression, we introduce an ε so small that 0 < θ − ε < θ < θ + ε < 1. From Lemma 2.1 we have

∫_{F_{m}} |T f_{n}| ≤ C_{±}2^{n+1}M_{0}^{1−θ∓ε}M_{1}^{θ±ε}μ(E_{n})^{1/p_{θ±ε}}ν(F_{m})^{1/q′_{θ±ε}}

for some constants C_{±} depending on q_{θ±ε}. Thus we have

∫_{Y} |T f||g| ≤ ∑_{n} ∑_{m} 2^{n+m+2} min_{±}(C_{±}M_{0}^{1−θ∓ε}M_{1}^{θ∓ε}μ(E_{n})^{1/p_{θ±ε}}ν(F_{m})^{1/q′_{θ±ε}})

where we use min_{±}(F_{±}) as short-hand for min(F_{−}, F_{+}). If we write

C′ := 4 max(C_{1}, C_{2}); a_{n} := 2^{np_{θ}}μ(E_{n}); b_{m} := 2^{mq′_{θ}}ν(F_{m})

then we have

∑_{n} a_{n}, ∑_{m} b_{m} ≤ 1

and

∫_{Y} |T f||g| ≤ C′M_{0}^{1−θ}M_{1}^{θ} ∑_{n} ∑_{m} min_{±}(2^{n(1− \frac{p_{θ}}{p_{θ±ε}})}2^{m(1− \frac{q′_{θ}}{q{′ \atop θ±ε}})}M_{0}^{∓ε}M_{1}^{±ε}a_{n}^{1/p_{θ±ε}}b_{m}^{1/q′_{θ±ε}}).

We can write the left-hand side as

C′M_{0}^{1−θ}M_{1}^{θ} ∑_{n} ∑_{m} min_{±}(A(n, m)^{±ε}a_{n}^{1/p_{θ±ε}}b_{m}^{1/q′_{θ±ε}})

where

A(n, m) = \frac{M_{1}}{M_{0}} 2^{−np_{θ}( \frac{1}{p_{0}} − \frac{1}{p_{1}} )}2^{mq′_{θ}( \frac{1}{q_{0}} − \frac{1}{q_{1}} )}.

Thus it will suffice to show that

∑_{n} ∑_{m} min_{±}(A(n, m)^{±ε}a_{n}^{1/p_{θ±ε}}b_{m}^{1/q′_{θ±ε}}) ≤ C′′

for some constant C′′. This may look like a mess, but it is a significant improvement, as we no longer see the original operator T or the functions f, g, and just have to deal with the sequences a_{n}, b_{m} and the very explicit function A(n, m).

We subdivide the above sum on the size of A(n, m), as

∑_{k} ∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} min_{±}(A(n, m)^{±ε}a_{n}^{1/p_{θ±ε}}b_{m}^{1/q′_{θ±ε}}).

Currently k can range over both positive and negative integers. It turns out that for positive k, the − sign is favourable, and for negative k the + sign is favourable. Thus it will suffice to show that

∑_{k=0}^{+∞} ∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} A(n, m)^{−ε}a_{n}^{1/p_{θ−ε}}b_{m}^{1/q′_{θ−ε}}) ≤ C′′_{−}

and

∑_{k=−∞}^{0} ∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} A(n, m)^{ε}a_{n}^{1/p_{θ+ε}}b_{m}^{1/q′_{θ+ε}}) ≤ C′′_{+}

for some constants C′′_{−}, C′′_{+}.

We shall just prove the first inequality; the second is similar and is left to the reader. Since 2^{k} ≤ A(n, m), we have A(n, m)^{−ε} ≤ 2^{−kε}. So we have

∑_{k=0}^{+∞} ∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} A(n, m)^{−ε}a_{n}^{1/p_{θ−ε}}b_{m}^{1/q′_{θ−ε}}) ≤ ∑_{k=0}^{+∞} 2^{−εk} ∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} a_{n}^{1/p_{θ−ε}}b_{m}^{1/q′_{θ−ε}}).

Since the geometric series ∑_{k=0}^{+∞} 2^{−εk} is convergent, it suffices to show that

∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} a_{n}^{1/p_{θ−ε}}b_{m}^{1/q′_{θ−ε}}) ≤ C′′′

for all k and some C′′′.

We use Schur’s test (and Hölder’s inequality). Observe that

∑_{n:2^{k}≤A(n,m)<2^{k+1}} 1 ≤ C′′′ for all m ∈ Z

and

∑_{m:2^{k}≤A(n,m)<2^{k+1}} 1 ≤ C′′′ for all n ∈ Z

for some C′′′, and hence

∑_{n,m:2^{k}≤A(n,m)<2^{k+1}} a_{n}^{1/p_{θ−ε}}b_{m}^{1/q′_{θ−ε}}) ≤ C′′′∥a^{1/p_{θ−ε}}∥_{l^{p_{θ−ε}}} ∥b^{1/q′_{θ−ε}}∥_{l^{p{′ \atop θ−ε}}} .

Now

∥a^{1/p_{θ−ε}}∥_{l^{p_{θ−ε}}} = (∑_{n} a_{n})^{1/p_{θ−ε}} ≤ 1.

A similar argument gives

∥b^{1/q′_{θ−ε}}∥_{l^{q{′ \atop θ−ε}}} ≤ 1.

But by hypothesis we have p_{0} ≤ q_{0} and p_{1} ≤ q_{1}, hence q′_{θ−ε} ≤ p′_{θ−ε}. For discrete measure spaces, lower l^{p} norms control higher ones, and hence

∥b^{1/q′_{θ−ε}}∥_{l^{p{′ \atop θ−ε}}} ≤ 1.

This proves the claim.

4. Wrapping up

Fix 0 < θ < 1. We are almost done. First, we observe from the monotone convergence theorem (writing an arbitrary |g| by the monotone limit of simple functions) that we can remove the hypothesis from Proposition 3.2 that g is simple, thus we have

∫_{Y} |T f||g| ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}∥g∥_{L^{q′_{θ}} (Y)}

for all f ∈ L^{p_{θ}}(X) and g ∈ L^{q′_{θ}(Y)}, with f simple. Applying duality we conclude that T f ∈ L^{q_{θ}}(Y) with

∥T f∥_{L^{q_{θ}} (Y)} ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}.

Now all we need to do is remove the hypothesis that f is simple. It is first convenient to truncate things a bit. Let A > 0 be an arbitrary number and let F ⊂ Y be a set of finite measure. Then we clearly have

∥ min(|T f|, A)∥_{L^{q_{θ}} (F)} ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}  (1)

for all simple f.

Now take f ∈ L^{p_{θ}}(X) which is not simple. We can write f as the limit in L^{p_{θ}}(X) of simple functions f^{(n)}, with |f^{(n)}| converging monotonically to |f| and f_{n} converging pointwise to f. Let us decompose f = f_{0} +f_{1}, where f_{0} ∈ L^{p_{0}}(X) and f_{1} ∈ L^{p_{1}}(X), by dividing X into the regions where f is large or small. This induces a similar decomposition f^{(n)} = f_{0}^{(n)} + f_{1}^{(n)}; from monotone convergence we can verify that f_{0}^{(n)} converges in L^{p_{0}} to f_{0} and similarly f_{1}^{(n)} converges in L^{p_{1}} to f_{1}. We then write by subadditivity and the triangle inequality

min(|T f|, A) ≤ min(|T f^{(n)}|, A) + min(|T (f_{0} − f_{0}^{(n)})|, A) + min(|T (f_{1} − f_{1}^{(n)})|, A).

Since f_{0} − f_{0}^{(n)} goes to zero in L^{p_{0}}(X) norm, the functions min(|T (f_{0} − f_{0}^{(n)})|, A) converge in weak L^{q_{0}}, and are also bounded. Hence they go to zero in L^{q′_{θ}}(F) norm (why? there are many ways to establish this. Note we will need F to have finite measure). Similarly min(|T (f_{1} − f_{1}^{(n)})|, A) goes to zero in L^{q′_{θ}}(F) norm. Taking L^{q′_{θ}}(F) norms of all sides, using (1) for f^{(n)}, and taking limits as n → ∞, we obtain

∥ min(|T f|, A)∥_{L^{q_{θ}} (F)} ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}.

LECTURE NOTES ON INTERPOLATION 7

Sending A to infinity using monotone convergence, we conclude

∥T f∥_{L^{q_{θ}} (F)} ≤ CM_{0}^{1−θ}M_{1}^{θ}∥f∥_{L^{p_{θ}} (X)}.

Sending F to infinity, using the hypothesis that Y is σ-finite and monotone convergence again, we obtain the claim.

Department of Mathematics, UCLA, Los Angeles CA 90095-1555

E-mail address: tao@math.ucla.edu